Question

求教定积分问题:积分限是0到2,求I=∫x(√2x-x^2)dx

是陈文灯的视频里面的一个练习题,他没写答案....自己弄不出来啊
积分限是0到2,求I=∫x(√2x-x^2)dx
也许我打的有点问题,是I=∫(0,2)x√(2x-x^2)dx

Answer 1

∫(0，2) x√(2x - x²) dx
= ∫(0，2) x√[- (x² - 2x + 1) + 1] dx
= ∫(0，2) x√[1 - (x - 1)²] dx
令x - 1 = sinθ，dx = cosθ dθ
x = 0 --> θ = - π/2
x = 2 --> θ = π/2
= ∫(- π/2，π/2) (1 + sinθ)|cosθ| * cosθ dθ
= ∫(- π/2，π/2) (1 + sinθ)cos²θ dθ
= ∫(- π/2，π/2) cos²θ dθ + ∫(- π/2，π/2) sinθcos²θ dθ
= 2∫(0，π/2) (1 + cos2θ)/2 dθ + ∫(- π/2，π/2) cos²θ d(- cosθ)
= [θ + (1/2)sin2θ] |(0，π/2) - (1/3)[cos³θ] |(- π/2，π/2)
= π/2

Answer 2

I=∫x(√2x-x^2)dx
=∫(√2x²-x³)dx
=∫√2x²dx-∫x³dx
=√2/3*x³-1/4*x^4+C
所以
∫<0,2>x(√2x-x^2)dx
=(√2/3*x³-1/4*x^4)|0,2
=8√2/3-4